1. **State the problem:** We need to find the values of $A$ and $B$ in the table for the function $$y = \frac{9}{2}x^2 + \frac{3}{2}x - 15$$ given some $x$ values and corresponding $y$ values. 2. **Recall the formula:** The function is quadratic: $$y = \frac{9}{2}x^2 + \frac{3}{2}x - 15$$ 3. **Calculate $A$ when $x = -2$:** $$y = \frac{9}{2}(-2)^2 + \frac{3}{2}(-2) - 15$$ $$= \frac{9}{2} \times 4 + \frac{3}{2} \times (-2) - 15$$ $$= 18 - 3 - 15$$ $$= 0$$ 4. **Calculate $B$ when $x = 1$:** $$y = \frac{9}{2}(1)^2 + \frac{3}{2}(1) - 15$$ $$= \frac{9}{2} + \frac{3}{2} - 15$$ $$= \frac{9 + 3}{2} - 15$$ $$= \frac{12}{2} - 15$$ $$= 6 - 15$$ $$= -9$$ 5. **Final answer:** $$A = 0, \quad B = -9$$